Answer
$$5t{e^t}\left( {2 + t} \right) - 6{t^2}{e^{ - t}}\left( {t - 3} \right)$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dt}}\left( {{t^2}\left( {{\bf{i}} + 2{\bf{j}} - 2t{\bf{k}}} \right) \cdot \left( {{e^t}{\bf{i}} + 2{e^t}{\bf{j}} - 3{e^{ - t}}{\bf{k}}} \right)} \right) \cr
& {\text{use the distributive property for }}{t^2}\left( {{\bf{i}} + 2{\bf{j}} - 2t{\bf{k}}} \right) \cr
& = \frac{d}{{dt}}\left( {\left( {{t^2}{\bf{i}} + 2{t^2}{\bf{j}} - 2{t^3}{\bf{k}}} \right) \cdot \left( {{e^t}{\bf{i}} + 2{e^t}{\bf{j}} - 3{e^{ - t}}{\bf{k}}} \right)} \right) \cr
& {\text{Calculate the dot product}} \cr
& = \frac{d}{{dt}}\left( {{t^2}{e^t} + 4{t^2}{e^t} + 6{t^3}{e^{ - t}}} \right) \cr
& = \frac{d}{{dt}}\left( {5{t^2}{e^t} + 6{t^3}{e^{ - t}}} \right) \cr
& {\text{Differentiate}} \cr
& = 10t{e^t} + 5{t^2}{e^t} + 18{t^2}{e^{ - t}} - 6{t^3}{e^{ - t}} \cr
& {\text{Factoring}} \cr
& = 5t{e^t}\left( {2 + t} \right) - 6{t^2}{e^{ - t}}\left( {t - 3} \right) \cr} $$
